Question

If you want to prepare 5 liters of a 0.35m solution of nh4cl, how many grams of salt

Answer 1

Answer is: mass fo ammonium chloride is 93.625 grams.
V(NH₄Cl) = 5 L.
c(NH₄Cl) = 0.35 M.
n(NH₄Cl) = V(NH₄Cl) · c(NH₄Cl).
n(NH₄Cl) = 5 L · 0.35 mol/L.
n(NH₄Cl) = 1.75 mol.
M(NH₄Cl) = 14 + 1·4 + 35.5 · g/mol = 53.5 g/mol.
m(NH₄Cl) = n(NH₄Cl) · M(NH₄Cl).
m(NH₄Cl) = 1.75 mol · 53.5 g/mol.
m(NH₄Cl) = 93.625 g.